Hardy Weinberg Problem Set 1 Answers
1. In fruit flies, the allele for normal wing length is dominant over the allele for vestigial wings
(these wings are stubby and therefore the flies are flightless). In a population of 1,000
individuals, 360 show the recessive phenotype. How many individuals would you expect to
be homozygous dominant and heterozygous for this trait?
p+q=1
and
p2 + 2(p)(q) + q2 = 1.0
Recessive phenotype: q2 = 360/1000 = .36
Homozygous Dominant: p2 (1000)
Recessive Allele: q = .6
p = 1 - .6
p = .4
.16(1000) = 160
Heterozygous: 2(p)(q) = 2(.6)(.4) = .48 (1000) = 480
2. The allele for the ability to roll your tongue is dominant over the allele for the lack of tongue
rolling. In a school like Hale, with a population of 1600 students, 25% show the recessive
phenotype. How many individuals would you expect to be homozygous dominant and
heterozygous for this trait?
p+q=1
and
p2 + 2(p)(q) + q2 = 1.0
Homozygous Dominant: p2 (1600)
q2 = 25% = .25
q = .5, p = .5
.52(1600) = 400
Heterozygous: 2(p)(q) = 2(.5)(.5) = .5 (1600) = 800
3. The allele for the hair pattern called “widow’s peak” is dominated over the allele for no
“widow’s peak”. Using the same school population as in question 2, how many individuals
would you expect for each of the three genotypes for the “widow’s peak” trait if 784
students have a widow’s peak.
p+q=1
and
p2 + 2(p)(q) + q2 = 1.0
Recessive phenotype: q2 = 784/1600 = .49
Homozygous Dominant: p2 (1600)
Recessive Allele: q = .7
p = 1 - .7
p = .3
.32(1600) = 144
Heterozygous: 2(p)(q) = 2(.7)(.3) = .42 (1600) = 672
4. In the United States, about 16% of the population is Rh negative. The allele for Rh
negative is recessive to the allele for Rh positive. If the student population for Hale is used,
how many students here at Hale would you expect in for each of the three possible
genotypes?
Recessive phenotype: q2 = 16%=.16
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AP BIOLOGY Krabath
Recessive Allele: q = .4
p = 1 - .4
p = .6
Homozygous Recessive Genotype: q2 (1600) =
Homozygous Dominate Genotype:
p2 (1600) =
.42(1600) = 256
.62(1600) = 576
Heterozygous Genotype : 2(p)(q) = 2(.4)(.6) = .48 (1600) = 768
5. In certain African Countries, 4% of newborn babies have sickle-cell anemia, which is a
recessive trait. Out of a random sample of 1,000 newborns, how many would you expect
for each of the three possible genotypes?
Recessive phenotype: q2 = 16%=.04
Recessive Allele: q = .2
Homozygous Recessive Genotype: q2 (1600) =
Homozygous Dominate Genotype:
p2 (1600) =
p = 1 - .4
p = .8
.22(1000) = 40
.82(1000) = 640
Heterozygous Genotype: 2(p)(q) = 2(.2)(.8) = . (1000) = 320
6. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time.
What is the frequency of the dominant allele?
Dominate Genotype includes the homozygous dominate and heterozygous which
occurs 91% of the time. There for the recessive genotype must show up 9% of the
time.
Recessive phenotype = q2 = .09 , q = .3
p = 1 - .3 p = .7
Dominant allele frequency is p which is .7 or 70%
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AP BIOLOGY Krabath